To find out when the rock hits the bottom of the crevasse, we need to set the function \( y = -4.9t^2 + 9 \) equal to \( -2 \) (since the bottom of the crevasse is 2 meters below ground level).
Setting up the equation:
\[ -4.9t^2 + 9 = -2 \]
Now, we can add 2 to both sides:
\[ -4.9t^2 + 11 = 0 \]
Next, we can isolate \( t^2 \):
\[ -4.9t^2 = -11 \]
Dividing by -4.9 gives:
\[ t^2 = \frac{11}{4.9} \]
Calculating \( \frac{11}{4.9} \):
\[ t^2 \approx 2.2449 \]
Taking the square root of both sides:
\[ t \approx \sqrt{2.2449} \approx 1.5 \]
So, rounding to the nearest tenth of a second, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.
The answer is: 1.5 seconds.
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